Sine Angle Product Formula 3

In this post, we continue the discussion from a previous post about the existence of a suitable sine angle product identity.

With the help of ChatGPT, I realized that I was missing an obvious solution here, based just on periodicity. In fact, we can show a much stronger statement: not only is there no polynomial or rational identity in terms of $\sin a,\cos a, \sin b,\cos b$ , there is no identity at all, period. $\sin(ab)$ simply does not depend on those four alone (without $a,b$ themselves.) More precisely:

Theorem. There is no function $f:\mathbb{R}^4\rightarrow\mathbb{R}$ such that for all real $a,b$ , $\sin(ab)=f(\sin a, \cos a, \sin b, \cos b)$ .

Proof. Assume such $f$ existed. Fix a value of $a$ , and define the one-variable function

$\displaystyle g(b) = \sin(ab) = f\left( \sin a,\cos a,\sin b,\cos b \right).$

For each $a$ , we have

$\displaystyle g(b + 2\pi) = f\left( \sin a,\cos a,\sin(b + 2\pi),\cos(b + 2\pi) \right)$

$\displaystyle = f\left( \sin a,\cos a,\sin b,\cos b \right)$

$\displaystyle = g(b),$

$\displaystyle \sin\left( a(b + 2\pi) \right) = \sin(ab)$

for all $a,b$ . But we clearly know this is false: either $a(b + 2\pi) = ab + 2\pi k$ for some integer $k$ or $a(b + 2\pi) = (2k + 1)\pi - ab$ for some integer $k$ , and there are real $a,b$ that do not satisfy either of these. The proof is complete. $\blacksquare$

This resolves the question I had been asking. A similar proof can be made for the analogous statement where angle multiplication is replaced by angle division too.

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